Question & Answer
QUESTION

Solve $cos\theta +\cos 2\theta +\cos 3\theta =0$.

ANSWER Verified Verified
Hint: Apply the identity of $\operatorname{cosC}+\cos D$ with the suitable terms of the given equation, which is given as $\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
Now, use the general solution of the equation$\cos x=coy$, after simplifying the given expression; which is given as:
If $\cos x=\cos y$
Then general solution is given as
$x=2n\pi \pm y$ Where $n\in z$

Complete step-by-step answer:
We have
$\cos \theta +\cos 2\theta +\cos 3\theta =0$ ………….. (i)
As we know the trigonometric identity of $\cos C+\cos D$ can be given as
$\operatorname{cosC}+cosD=2cos\dfrac{C+D}{2}\cos \dfrac{C-D}{2}$ …………… (ii)
Now, we can observe the identity (ii) and get that if C + D and C - D will be even numbers then angles in the cosine functions will be of integer type otherwise, we get angles of them in fractions.
So, now we can observe equation (i) and get that we should add $\cos \theta ,\cos 3\theta $with the help of property (ii) as sum of $\theta ,3\theta $ will be an even number. So, we can write equation (i) as:
$\left( \cos \theta +\cos 3\theta \right)+\cos 2\theta =0$
Apply the identity in equation (ii) with the first two terms of the above equation. So, we get
$2\cos \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)+\cos 2\theta =0$
$2\cos \left( 2\theta \right)\cos \left( -\theta \right)+\cos 2\theta =0$
We know that$\cos \left( -\theta \right)=\cos \theta $. So, we get
$2\cos \theta \cos \theta +\cos 2\theta =0$
Now, take $'\cos 2\theta '$ as common from both the terms of the above equation and hence, we get
$\cos 2\theta \left( 2\cos \theta +1 \right)=0$ ………………(iii)
As we know that multiplication of two terms will be zero, if one of them will be 0 or both. So, we get from equation (iii) as
$\cos 2\theta =0,2\cos \theta +1=0$
$\cos 2\theta =0,\cos \theta =\dfrac{-1}{2}$…………… (iv)
We know that general solution of equation $\cos x=\cos y$ is given by relation
$x=2n\pi \pm y$ ………… (v)
Where $n\in z$(integer)
So, we have
$\cos 2\theta =0$
As we know $\cos \theta $ gives value ‘0’ at $\theta =\dfrac{\pi }{2}$ , hence we get
$\cos 2\theta =\cos \dfrac{\pi }{2}$
Now, using relation (v), we get $\theta $ as
$2\theta =2n\pi \pm \dfrac{\pi }{2}$
Divide the whole equation by 2, we get
$\dfrac{2\theta }{2}=\dfrac{2n\pi }{2}\pm \dfrac{\pi }{4}$
$\theta =n\pi \pm \dfrac{\pi }{4}$………………… (vi)
And we have another relation as
$\cos \theta =\dfrac{-1}{2}$
As we know that $\cos \theta $ will give $\dfrac{1}{2}$ at $\theta =\dfrac{\pi }{3}$and
We know the relation$\cos \left( 180-\theta \right)=-\cos \theta $. So, we can put $\theta =\dfrac{\pi }{3}$ here as well and hence, we get
$\begin{align}
  & \cos \left( 180-\theta \right)=\cos \left( \pi -\theta \right)=-\cos \theta \\
 & \cos \left( \pi -\dfrac{\pi }{3} \right)=-\cos \dfrac{\pi }{3}=\dfrac{-1}{2} \\
\end{align}$
Or
$\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}$
It means $\cos \theta $ will give ‘$\dfrac{-1}{2}$’ at $\theta \dfrac{2\pi }{3}$
Hence, we can write the relation $\cos \theta =\dfrac{-1}{2}$ as $\cos \theta =\cos \dfrac{2\pi }{3}$
Hence, from the relation (v), we get
$\theta =2n\pi \pm \dfrac{2\pi }{3}$ …………….. (vii)
Hence, we get the solution of equation given in the problem $\left( \cos \theta +\cos 2\theta +\cos 3\theta =0 \right)$ as
$\theta =n\pi \pm \dfrac{\pi }{4},\theta =2n\pi \pm \dfrac{2\pi }{3}$

Note: One may go wrong if he/she uses the given identity i.e. $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ with first two terms or the last two terms i.e. $\cos \theta +\cos 2\theta ,\cos 2\theta +\cos 3\theta $. As we will get a fraction from $\theta $ inside the cosine functions and which will make the given relation complex. It means observing the given relations id the key point of the question.
One may go wrong if he/she solve the question in the following approach:
$\begin{align}
  & \cos \theta +\cos 3\theta =-\cos 2\theta \\
 & 2\cos 2\theta \cos \theta =-\cos 2\theta \\
\end{align}$
Cancelling the terms $\cos 2\theta $ from both sides, we get $2\cos 2\theta =-1$
So, he/she will miss the solution of the equations$\cos 2\theta =0$. So, take care of it and don’t cancel out the terms $\cos 2\theta $ from both sides as the number of solutions will become less.